最佳化問題建模與求解技術探討

from ortools.constraint_solver import routing_enums_pb2
from ortools.constraint_solver import pywrapcp

def create_data_model():
    """Stores the data for the problem."""
    data = {}
    data['distance_matrix'] = [
        [0, 10, 15, 20],
        [10, 0, 35, 25],
        [15, 35, 0, 30],
        [20, 25, 30, 0]
    ]
    data['num_vehicles'] = 1
    data['depot'] = 0
    return data

def print_solution(manager, routing, solution):
    """Prints solution on console."""
    print('Objective: {}'.format(solution.ObjectiveValue()))
    index = routing.Start(0)
    plan_output = 'Route for vehicle 0:\n'
    route_distance = 0
    while not routing.IsEnd(index):
        plan_output += ' {} ->'.format(manager.IndexToNode(index))
        previous_index = index
        index = solution.Value(routing.NextVar(index))
        route_distance += routing.GetArcCostForVehicle(previous_index, index, 0)
    plan_output += ' {}\n'.format(manager.IndexToNode(index))
    plan_output += 'Route distance: {}\n'.format(route_distance)
    print(plan_output)

def main():
    """Entry point of the program."""
    # Instantiate the data problem.
    data = create_data_model()

    # Create the routing index manager.
    manager = pywrapcp.RoutingIndexManager(len(data['distance_matrix']), 
                                           data['num_vehicles'], 
                                           data['depot'])

    # Create Routing Model.
    routing = pywrapcp.RoutingModel(manager)

    def distance_callback(from_index, to_index):
        """Returns the distance between the two nodes."""
        from_node = manager.IndexToNode(from_index)
        to_node = manager.IndexToNode(to_index)
        return data['distance_matrix'][from_node][to_node]

    transit_callback_index = routing.RegisterTransitCallback(distance_callback)

    # Define cost of each arc.
    routing.SetArcCostEvaluatorOfAllVehicles(transit_callback_index)

    # Setting first solution heuristic.
    search_parameters = pywrapcp.DefaultRoutingSearchParameters()
    search_parameters.first_solution_strategy = (
        routing_enums_pb2.FirstSolutionStrategy.PATH_CHEAPEST_ARC)

    # Solve the problem.
    solution = routing.SolveWithParameters(search_parameters)

    # Print solution on console.
    if solution:
        print_solution(manager, routing, solution)

if __name__ == '__main__':
    main()

# 定義決策變數
x0 = 0  # 蟾蜍數量
x1 = 0  # 蠑螈數量
x2 = 0  # 蚓螈數量

# 定義約束條件
def constraint_worms(x0, x1, x2):
    return 2*x0 + x1 + x2 <= 1500

def constraint_crickets(x0, x1, x2):
    return x0 + 3*x1 + 2*x2 <= 3000

def constraint_flies(x0, x1, x2):
    return x0 + 2*x1 + 3*x2 <= 5500

# 定義目標函式
def objective(x0, x1, x2):
    return x0 + x1 + x2

from ortools.linear_solver import pywraplp

def solve_coexistence():
    t = 'Amphibian coexistence'
    s = pywraplp.Solver(t, pywraplp.Solver.GLOP_LINEAR_PROGRAMMING)
    x = [s.NumVar(0, 1000, 'x[%i]' % i) for i in range(3)]
    pop = s.NumVar(0, 3000, 'pop')
    s.Add(2*x[0] + x[1] + x[2] <= 1500)
    s.Add(x[0] + 3*x[1] + 2*x[2] <= 3000)
    s.Add(x[0] + 2*x[1] + 3*x[2] <= 4000)
    s.Add(pop == x[0] + x[1] + x[2])
    s.Maximize(pop)
    s.Solve()
    return pop.SolutionValue(), [e.SolutionValue() for e in x]

# 範例程式碼：定義一個簡單的線性連續模型
from ortools.linearsolver import pywraplp

def solve_linear_continuous_model():
    # 建立一個線性規劃求解器
    solver = pywraplp.Solver.CreateSolver('GLOP')

    # 宣告變數
    x = [solver.NumVar(0, solver.infinity(), f'x{i}') for i in range(3)]

    # 新增約束條件
    constraint1 = solver.Constraint(0, 1500)
    constraint1.SetCoefficient(x[0], 2)
    constraint1.SetCoefficient(x[1], 1)
    constraint1.SetCoefficient(x[2], 1)

    # 設定目標函式
    objective = solver.Objective()
    for var in x:
        objective.SetCoefficient(var, 1)
    objective.SetMaximization()

    # 求解模型
    status = solver.Solve()

    # 傳回結果
    if status == pywraplp.Solver.OPTIMAL:
        return solver.Objective().Value(), [var.solution_value() for var in x]
    else:
        return None, None

#### 內容解密：
此範例程式碼展示瞭如何使用Google的OR-Tools函式庫建立和求解一個簡單的線性連續模型。首先，我們建立了一個線性規劃求解器例項，並聲明瞭一組連續變數。接著，我們增加了一個線性約束條件，並定義了一個簡單的線性目標函式，旨在最大化變數的總和。最後，我們呼叫求解器來計算最優解，並傳回目標函式的最優值和決策變數的最優值。

#### 進一步解析：
1. **建立求解器**：我們使用`pywraplp.Solver.CreateSolver('GLOP')`建立了一個根據GLOP後端的線性規劃求解器。GLOP是Google開發的一個高效的線性規劃求解器。
2. **變數宣告**：透過`solver.NumVar`方法，我們聲明瞭一組連續變數$x_0, x_1, x_2$，每個變數的取值範圍從0到無窮大。
3. **約束條件**：我們建立了一個約束條件$2x_0 + x_1 + x_2 \leq 1500$，並透過`SetCoefficient`方法設定了變數的係數。
4. **目標函式**：目標函式被設定為最大化$x_0 + x_1 + x_2$，體現了線性目標函式的特性。
5. **求解與結果**：呼叫`solver.Solve()`進行求解後，我們檢查瞭解的狀態。如果狀態為`OPTIMAL`，則傳回最優目標函式值和最優變數值。

### 線性連續模型的優勢

線性連續模型由於其簡單性和成熟的求解演算法（如單純形法和內點法），在眾多領域得到了廣泛應用。它們特別適合於資源分配、生產排程、投資組合最佳化等問題。